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3 years ago

Why does the frequency of a wave decrease as the wavelength increases?

1 answer:

6 0

Because the product of (wavelength) times (frequency) is always the same number ... the wave speed. Since their product can't change, then obviously if one of them changes, the other one must change in the opposite direction.

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a sample of iron has the dimensions of 2cm times 3cm times 2cm. if the mass of this rectangular-shaped object is 94g, what is th

Anarel [89]

The answer is 7.83. The three is repeating.

5 0

3 years ago

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

Pie

Answer:

a) v = 3,843 m / s, b) 46.7º North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East

initial instant. Before the crash

p₀ = mₐ vₐ₀

final insttne. After the crash

p_f = (mₐ + m_b) vₓ

p₀ = p_f

mₐ vₐ₀ = (mₐ + m_b) vₓ

vₓ = $\frac{m_a}{m_a + m_b} \ v_{ao}$

let's calculate

vₓ = $\frac{16.7}{16.7 + 29.3} \ 7.26$

vₓ = 2,636 m / s

direction north

initial p₀ = m_b v_{bo}

final p_f = (mₐ + m_b) v_y

p₀ = p_f

m_b v_{bo} = (mₐ + m_b) v_y

v_y = $\frac{m_b}{m_a+m_b} \ v_{bo}$

let's calculate

v_y = $\frac{29.3}{16.7 + 29.3} \ 4.39$

v_y = 2.796 m / s

the final speed of the two two vehicles is

v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{2.636^2 + 2.796^2}$

v = 3,843 m / s

b) let's use trigonometry to find the direction

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ (2,796 / 2,636)

θ = 46.7º

This direction is 46.7º North East

7 0

3 years ago

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

How is light generated?

Fudgin [204]

<span>electrons change momentum, some of them slough off photons. And some of those photons have energy in the visible light range of the electro-magnetic spectrum. </span>

5 0

3 years ago

Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the vel

DENIUS [597]

Answer:

28.3 m/s

Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

Vf -12.5 = 0.45(35)

Vf= 28.3 m/s

5 0

3 years ago